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Suppose I have two Cartan geometries $(\mathscr{G}_1,\omega_1)$ and $(\mathscr{G}_2,\omega_2)$ of type $(G,H)$ over the same manifold $M$. What conditions on $G$ and $H$ allow us to conclude that $\mathscr{G}_1$ and $\mathscr{G}_2$ are isomorphic as principal $H$-bundles?

It seems to be a common implicit assumption in the literature that $\mathscr{G}_1$ is always isomorphic to $\mathscr{G}_2$ in the cases we usually look at. In particular, for parabolic geometries, it seems to be folklore that this is true.

Previously, I had implicitly assumed that such an isomorphism always exists for Cartan geometries of all types, but I recently thought of the following example. If I have a Hermitian holomorphic line bundle, then I can construct a Cartan geometry of type $(\mathbb{C}^m\rtimes\mathrm{U}(1),\mathrm{U}(1))$ corresponding to the Chern connection. However, in general, there are too many line bundles over a given complex manifold for them to all be associated (in the sense that $L\cong\mathscr{G}\times_{\mathrm{U}(1)}\mathbb{C}$) to the same principal $\mathrm{U}(1)$-bundle, so there must be nonisomorphic principal $\mathrm{U}(1)$-bundles admitting Cartan connections of this type over the same manifold.

I’ve thought about this for a few days now, and I imagine there’s probably a nice general condition on $(G,H)$, but I’m not seeing what that condition might be.

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    $\begingroup$ I had examples in one of my papers of non-isomorphic holomorphic bundles arising as total spaces of holomorphic Cartan geometries. I think I proved that this doesn't happen for holomorphic parabolic geometries. But I don't know about real geometries. $\endgroup$
    – Ben McKay
    Commented Sep 12, 2020 at 7:56
  • $\begingroup$ In fact, I have a simple example of non-isomorphic holomorphic bundles arising as total spaces of holomorphic Cartan geometries, with the same model, on complex surfaces: arxiv.org/abs/1105.4732, but they are isomorphic as real bundles. $\endgroup$
    – Ben McKay
    Commented Jul 3, 2021 at 8:09

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This is not a complete answer, but I think that it might help to clear up some misunderstandings. It is not, in general, true that all the principal $H$-bundles over $M$ supporting a Cartan connection of type $(G,H)$ are isomorphic, though the OP's proposed example does not actually show this. I think that the following discussion may help.

To fix notation, let's recall what we mean by a "Cartan connection of type $(G,H)$": Here $G$ is a Lie group with Lie algebra $\frak{g}$ and $H$ is a Lie subgroup with Lie algebra ${\frak{h}}\subset{\frak{g}}$. The representation $\mathrm{Ad}:H\to\mathrm{Aut}({\frak{g}})$ preserves the subalgebra ${\frak{h}}$ and so induces a representation $\rho:H\to \mathrm{Aut}({\frak{g/h}})$. If $\pi:B\to M$ is a principal right $H$-bundle, let $X_v$ for $v\in\frak{h}$ be the vertical vector field on $B$ whose flow is right action by $\mathrm{exp}(tv)$. Then a Cartan connection of type $(G,H)$ on $\pi:B\to M$ is a $\frak{g}$-valued $1$-form $\gamma:TB\to \frak{g}$ with the following properties:

  1. $\gamma_u:T_uB\to{\frak{g}}$ is an isomorphism for all $u\in B$.
  2. $\gamma\bigl(X_v(u)\bigr) = v$ for all $u\in B$ and all $v\in\frak{h}$.
  3. $R^*_h(\gamma) = \mathrm{Ad}(h^{-1})(\gamma)$ for all $h\in H$.

It is important to note that not every principal right $H$-bundle over $M$ supports a Cartan connection of type $(G,H)$. This is because such a Cartan connection $\gamma$ defines an isomorphism $\iota_\gamma:TM\to B\times_\rho {\frak{g/h}}$. To see this, let $\omega = \gamma\,\mathrm{mod}\,{\frak{h}}:TB\to {\frak{g/h}}$. The above axioms imply that $\omega_u:T_uB/V_uB\to {\frak{g/h}}$ is an isomorphism for all $u\in B$, where $V_uB\subset T_uB$ is tangent to the fiber of $\pi:B\to M$. Since we have a canonical isomorphism $T_uB/V_uB\to T_{\pi(u)}M$, it follows that we can regard $\omega$ as defining an isomorphism $\omega_u:T_{\pi(u)}M\to {\frak{g/h}}$ for all $u\in B$ that satisfies $\omega_{u\cdot h} = \rho(h^{-1})(\omega_u)$ for all $u\in B$ and all $h\in H$. By the very definition of $B\times_\rho{\frak g/h}$, this establishes the claimed isomorphism $\iota_\gamma:TM\to B\times_\rho{\frak g/h}$.

Conversely, if an isomorphism $\iota:TM\to B\times_\rho{\frak g/h}$ is given, then one can construct a Cartan connection of type $(G,H)$ on $B$.

Thus, one can see why the OPs construction starting with a line bundle $L$ endowed with a $\mathrm{U}(1)$-connection does not automatically imply that there is a Cartan connection of the desired type on $M$. For example, in this case, if a Cartan connection existed, then $TM$ would have to be isomorphic to $L\otimes \mathbb{C}^n = B\times_\rho {\frak g/h}$, and this is generally not the case.

However, there is a simpler example to demonstrate that not all $H$-bundles that admit Cartan connections of type $(G,H)$ are isomorphic: Here, let $n=3$, let $H=\mathrm{SO}(2)$ and let $G = \mathbb{R}^3\rtimes H$, where $H=\mathrm{SO}(2)$ acts on $\mathbb{R}^3$ by rotation in the second and third coordinates. An $H$-bundle $\pi:B\to M^3$ is just an $\mathrm{SO}(2)$-bundle, so it has an Euler class (which could be nonzero) and the associated bundle $B\times_\rho \mathbb{R}^3$ is a sum of a trivial bundle and a $2$-plane bundle. If there is a Cartan connection on $B$, then we get an isomorphism of $TM$ with the sum of a trivial bundle and a $2$-plane bundle. In particular, this means that $M$ is oriented and we have a nonvanishing vector field on $M$ together with a $2$-plane subbundle that has a well-defined Euler class.

Now, every oriented $3$-manifold has a trivial tangent bundle, but once one chooses a nonvanishing vector field, the Euler class of the complementary $2$-plane bundle is determined and may very well be nonzero. For example, let $M = S^1\times S^2$. If we choose the vector field tangent to the $S^1$-fibers, then the complementary $2$-plane field is nontrivial on each $S^2$-fiber. Meanwhile, if we choose a trivialization of the tangent bundle of $M$, then letting the vector field be one of the three trivializing vector fields, the complementary $2$-plane bundle will be trivial.

Thus, we can have two $H$-bundles over $M$ that are not isomorphic even though they both admit Cartan connections of type $(\mathbb{R}^3\rtimes H,\ H)$.

It follows that the very first criterion one needs to have in order for all the Cartan connections of type $(G,H)$ to have isomorphic underlying $H$-bundles is that all of the structure reductions of the canonical $\mathrm{GL}(n,\mathbb{R})$-structure on $TM$ to a $\rho(H)$-structure be isomorphic. This is a very strong condition on $\rho(H)$ and $M$, and whether it is met depends on both $\rho(H)$ and $M$.

Meanwhile, for most of the familiar examples in parabolic geometry, $\rho(H)$ is some large group such as $\mathrm{GL}(n,\mathbb{R})$, $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{CO}(n)$, or $\mathrm{SO}(n)$, and it happens that this uniqueness is met trivially. This may account for the common (false) belief that prompted this question in the first place.

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  • $\begingroup$ Thank you so much. I’m kind of embarrassed that I convinced myself Chern connections induce Cartan connections, but your post was helpful in clearing up why I was wrong and (partially) answering the main question. $\endgroup$ Commented Sep 21, 2020 at 13:50
  • $\begingroup$ @RobinGoodfellow: Well, I now think that, probably, the answer to the main question is going to be that "$\rho(H)$ contains a maximal compact subgroup of $\mathrm{Aut}(\frak{g/h})$" is the necessary and sufficient condition that all the $H$-bundles over any $M$ that support a Cartan connection of type $(G,H)$ be isomorphic. $\endgroup$ Commented Sep 21, 2020 at 14:01
  • $\begingroup$ As a small follow-up: you say that most familiar examples in parabolic geometry satisfy your criterion trivially. While I agree for the usual examples, for more general parabolic geometries this seems like it might be questionable. For example, if I have a parabolic geometry from the (adjoint) split-real form of the exceptional simple Lie group of rank 2 coming from crossing out the smaller node of the Satake diagram, then $G_0$ will be isomorphic to $\mathrm{GL}_2\mathbb{R}$, which feels too small to trivially satisfy the criterion... $\endgroup$ Commented Sep 21, 2020 at 14:04
  • $\begingroup$ ...Is it known whether it holds for all parabolic geometries? $\endgroup$ Commented Sep 21, 2020 at 14:05
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    $\begingroup$ @RobinGoodfellow: Well, yes. There will be such examples for $m=3$: Just take the two non-isomorphic examples for $\bigl(\mathbb{R}^3\rtimes \mathrm{SO}(2), \mathrm{SO}(2)\bigr)$ on $3$-manifolds that I mentioned and notice that $\rho\bigl(\mathrm{SO}(2)\bigr)$ sits as a maximal compact in $\mathrm{SO}^\circ(1,2)\subset\mathrm{GL}(3,\mathbb{R})$. In other words, there will be $3$-manifolds (e.g. $M=S^1\times S^2$) that admit two oriented, time-oriented conformal Lorentzian structures whose underlying Cartan conformal connection bundles are not isomorphic. $\endgroup$ Commented Sep 21, 2020 at 15:42

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